a) Ta có : 51ⁿ=\(\overline{...1}\)

                47¹⁰²=47².(47⁴)²⁵=\(\left(\overline{...9}\right).\left(\overline{...1}\right)=\overline{...9}\)

\(\Rightarrow51^n+47^{102}=\left(\overline{...1}\right)+\left(\overline{...9}\right)=\overline{...0}⋮10\)

Vậy 51ⁿ+47¹⁰²\(⋮\)10.

b) Ta có : \(17^5=17.17^4=17.\left(\overline{...1}\right)=\overline{...7}\)

                \(24^4=\overline{...6}\)

                 \(13^{21}=13.\left(13^4\right)^5=13.\left(\overline{...1}\right)=\overline{...3}\)

\(\Rightarrow17^5+24^4-13^{21}=\left(\overline{...7}\right)+\left(\overline{...6}\right)-\left(\overline{...3}\right)=\overline{...0}⋮10\)

Vậy 17⁵+24⁴+13²¹\(⋮\)10

a) 17²⁰⁰⁸  = (17⁴)⁵⁰² = (...1)⁵⁰² = (....1)

11²⁰⁰⁸ = (....1)

3²⁰⁰⁸ = (3⁴)⁵⁰² = (...1)⁵⁰²  = (...1)

=> 17²⁰⁰⁸ - 11²⁰⁰⁸ - 3²⁰⁰⁸ = (...1) - (...1) - (...1)

Hiệu 17²⁰⁰⁸ - 11²⁰⁰⁸ tận cùng là 0 => 17²⁰⁰⁸ - 11²⁰⁰⁸ - 3²⁰⁰⁸ tận cùng là 9

b) 17²⁵ = (17⁴)⁶.17 = (...1)⁶.17 = (...7)

24⁴ = (24²)² = (...6)² = (...6)

13²¹ = (13⁴)⁵.13 = (...1)⁵.13 = (...3)

=> B = 17²⁵ - 24⁴ - 13²¹ = (...7) + (...6) - (....3) = (....0) => B chia hết cho 10

c) Tương tự

ta co 17^5+24^4-13^21

=17^4.17+...6-13^(4.5).13

=...7+...6-...3

=...0

do so tan cung la 0 nen chia het cho 10 =)))